Describe the effect of surface colour (black or white) and texture (dull or shiny) on the emission, absorption and reflection of infrared radiation

2.3.3 Radiation – Colour & Texture Effects

Key Concepts

• Absorptivity (\$α\$): fraction of incident IR energy absorbed.
• Reflectivity (\$ρ\$): fraction of incident IR energy reflected.
• Emissivity (\$ε\$): fraction of thermal radiation emitted by a surface.
• For any surface, \$α + ρ + τ = 1\$, where \$τ\$ is transmissivity (usually zero for opaque solids).
• Black surfaces have \$α ≈ 1\$, \$ρ ≈ 0\$, \$ε ≈ 1\$.
• White surfaces have \$α ≈ 0\$, \$ρ ≈ 1\$, \$ε ≈ 0\$.

Colour Matters – Black vs White

Think of a black hoodie on a sunny day. It feels hot because it absorbs most of the IR from the sun. A white hoodie feels cooler because it reflects IR. This is the same principle for infrared radiation from any source.

Texture Matters – Dull vs Shiny

A shiny metal surface (like a polished spoon) reflects IR like a mirror, so \$ρ\$ is high. A dull surface (like a matte paint) scatters IR, reducing \$ρ\$ and increasing \$α\$ slightly. The difference is subtle but important for heat transfer.

Summary Table

Exam Tips 💡

Remember: In exam questions, colour and texture are often used to test your understanding of \$α\$, \$ρ\$, and \$ε\$. When a surface is described as “black and dull”, you can safely assume \$α≈1\$, \$ρ≈0\$, \$ε≈1\$. For “white and shiny”, think \$α≈0\$, \$ρ≈1\$, \$ε≈0\$ (unless the problem specifies otherwise). Use the equation \$α + ρ + τ = 1\$ to check your reasoning. Good luck! 🌟